The enthalpy of formation of ethane and benzene from the gaseous atoms are $- 2839.2 k J / m o l$ and $- 5506 k J / m o l$ respectively. Bond enthalpy of $C = C$ bond is
Given: Resonance energy of benzene $= - 23.68 k J / m o l$
Bond enthalpy of $C - H$ bond $= 411.0 k J / m o l$

Question

The enthalpy of formation of ethane and benzene from the gaseous atoms are $- 2839.2 k J / m o l$ and $- 5506 k J / m o l$ respectively. Bond enthalpy of $C = C$ bond is

Given: Resonance energy of benzene $= - 23.68 k J / m o l$

Bond enthalpy of $C - H$ bond $= 411.0 k J / m o l$

TestHub · Accepted Answer

− 2839.2 = 0 − [ 6 × 411 + Δ H C − C ​ ] Δ H C − C ​ = 373.2 k J / m o l 6 C ( g ) + 6 H ( g ) ⟶ C 6 ​ H 6 ​ ( g ) Δ H theoritical ​ = 0 − [ 6 × 411 + 3 × 373.2 + 3 x ] R.E. = Actual - Theoritical − 23.68 = − 5506 + [ 6 × 411 + 3 × 373.2 + 3 x ] 3 x = 5506 − 23.68 − 2466 − 1119.6 x = 632.24 k J / m o l or Δ H C = C ​ = 632.24 k J / m o l ]

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